Prove log(n)/n^sigma is a null sequence.

[elliti123]

Mod note: Moved from the homework section.
1. Homework Statement

If σ > 0 and base b > 1 then prove that $log n/n^σ$ is a null sequence. This is not really a homework since i am self studying Konrad Knopp book about infinite series and i wanted to see different ideas and perspectives on the proof of the problem.Thanks.

 

[jedishrfu]

Can you provide your proof of the problem?

Also what is the b value? Is it the base of the log function?

 

[elliti123]

Yes the b is the base of the log function.

 

[elliti123]

Mod note: Edited text below to make it more readable.
If b > 1 then we have $b^σ > 1$. Therefore $(n/(b^σ)^n)$ is a null sequence.
Given ε > 0 we have consequently from a point onwards - say for every n > m, $(n/(b^σ)^n) < ε' = ε/b^σ$. But in any case, $log n/n^σ < b^σ(g + 1/(b^σ)^g b^1 ))$ . If g denotes the characteristic of log n . If therefore, $n > b^m$ Then, $|(log\ n/n^σ)| < ε ∀\ n > b^m$ Q.E.D

This is the proof of the book and it is a theorem not in the exercises as the author is building up the knowledge of the reader, but as i read this again i have one question now how did he come up with what n has to be bigger than ?